P-Values for Hypothesis Testing About  With  Known.

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p-Values for Hypothesis Testing About  With  Known

Hypothesis Testing (Revisited) Five Step Procedure 1.Define Opposing Hypotheses. (  ) 2.Choose a level of risk (  ) for making the mistake of concluding something is true when its not. 3.Set up test (Define Rejection Region). random sample 4.Take a random sample. 5.Calculate statistics and draw a conclusion.

Concept of a p-value Ignore step 3 Calculate The probability of getting an value at least as far away as the observed if H 0 were true value, if H 0 were true. p-Value

Calculating p-Values A p-value is a probability whose definition varies depending on the type of test we are doing (i.e. the form of the alternate hypothesis.) Alternate Hypothesis p-value “>”Area to the right of “<”Area to the left of “  ”2* Area in the “tail” (to the right or left of )

25 p-value For “> Tests” = P(Getting a value greater than When H 0 is true) H 0 :  = 25 H A :  > 25 p-value =  = 4.2, n = 49 0 Z z = (26.23-25)/.6 2.05.9798 p = 1-.9798.0202 =.0202

27 p-value For “< Tests” = P(Getting a value less than When H 0 is true) H 0 :  = 27 H A :  < 27 p-value =  = 4.2, n = 49 0 Z z = (26.23-27)/.6 -1.28.1003 p =.1003

26 p-value For “  Tests” = P(Getting a value at least as far away as When H 0 is true) H 0 :  = 26 H A :   26  = 4.2, n = 49 0 Z.38.23 below 26.23 above 26 z = (26.23-26)/.6 p-value = Area above 26.23 + Area below 25.77 = 2*Area above 26.23.6480.3520 p = 2(.3520).7040 =.7040

P-VALUES AND α Consider H A :  > 25 – Here we got z = 2.05 –Since  =.05, z.05 = 1.645 Can Accept H A –Suppose  =.01; z.01 = 2.326 Cannot Accept H A –What about  =.02? z.02 = 2.054 Cannot Accept H A –What about  =.03? z.03 = 1.88 Can Accept H A p-valueThere is some value of  that is the “break-point” between accepting and not accepting H A -- this is the p-value. Accept H A If p  α, Accept H A Do Not Accept H A If p > α, Do Not Accept H A LOW p-values are SIGNIFICANT!!

=AVERAGE(A2:A50) =(C6-C3)/(C2/SQRT(49)) =1 – NORMSDIST(C7)

=AVERAGE(A2:A50) =(C6-C3)/(C2/SQRT(49)) =NORMSDIST(C7)

=AVERAGE(A2:A50) =(C6-C3)/(C2/SQRT(49)) =2*(1-NORMSDIST(C7)) Note: If z were negative, the p-value would have been: =2*NORMSDIST(C7)

REVIEW p-values measure the strength of the test –lower p-values indicate more strongly that H A is true p-values –“>” tests -- Area in upper tail (to the right of ) –“<” tests -- Area in lower tail (to the left of ) –“  ” tests -- twice the area in a “tail” If z >0 -- twice the area in the upper tail If z< 0 -- twice the area in the lower tail